Consider a matrix $$A = \begin{pmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{pmatrix}$$, where $$\alpha, \beta, \gamma$$ are three distinct natural numbers. If $$\frac{\det(\text{adj}(\text{adj}(\text{adj}(\text{adj} A))))}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}} = 2^{32} \times 3^{16}$$, then the number of such 3-tuples $$(\alpha, \beta, \gamma)$$ is ______.

For an $$n \times n$$ matrix $$M$$ we recall three standard facts:
  1. $$M\,\text{adj}\,M = \det M \, I_n$$(definition of adjugate).
  2. $$\det(\text{adj}\,M)= (\det M)^{\,n-1}\;. $$
  3. $$\text{adj}(\text{adj}\,M)= (\det M)^{\,n-2}\,M\;.$$

In our problem $$n=3$$, so $$\text{adj}(\text{adj}\,M)= (\det M)\,M\;.$$

Put $$A_0=A,\;A_1=\text{adj}\,A,\;A_2=\text{adj}\,A_1,\;A_3=\text{adj}\,A_2,\;A_4=\text{adj}\,A_3\;.$$ Let $$D=\det A\;.$$ Working step-by-step:

Case 1: $$A_1=\text{adj}\,A \;\Longrightarrow\; \det A_1=D^{2}\;.$$

Case 2: $$A_2=\text{adj}(A_1)= (\det A)\,A = D\,A\;.$$

Case 3: $$A_3=\text{adj}(A_2)=\text{adj}(D\,A)=D^{2}\,\text{adj}A=D^{2}A_1\;.$$

Case 4: $$A_4=\text{adj}(A_3)=\text{adj}(D^{2}A_1)=D^{4}\,\text{adj}A_1=D^{4}A_2=D^{4}\,(D\,A)=D^{5}A\;.$$

Since a scalar multiple rescales the determinant by the cube of that scalar (for a $$3\times3$$ matrix), we get

$$\det A_4=(D^{5})^{3}\,D=D^{15}\,D=D^{16}\;.$$

Therefore $$\det\bigl(\text{adj}(\text{adj}(\text{adj}(\text{adj}A)))\bigr)= (\det A)^{16}\;.$$

The question supplies

$$\frac{(\det A)^{16}}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}}=2^{32}\times3^{16}\;.$$

Hence we must find $$\det A$$. Writing out

$$A=\begin{pmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{pmatrix},$$

expand the determinant once (routine algebra gives)

$$\det A=(\beta-\alpha)\gamma^{3}+(\gamma-\beta)\alpha^{3}+(\alpha-\gamma)\beta^{3}\;.$$

This expression is anti-symmetric and factors exactly like the Vandermonde determinant:

$$\det A=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)\,(\alpha+\beta+\gamma)\;.$$

Substituting into the given relation we have

$$\bigl[(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)\bigr]^{16} \;\Big/ \; (\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16} \;=\;(\alpha+\beta+\gamma)^{16}=2^{32}\times3^{16}\;.$$

Because $$2^{32}\times3^{16}=(2^{2}\times3)^{16}=12^{16},$$ we obtain

$$\alpha+\beta+\gamma=12.$$

Finally, $$\alpha,\beta,\gamma$$ are distinct natural numbers. List unordered triples $$a\lt b\lt c$$ with $$a+b+c=12$$:

$$\{1,2,9\},\{1,3,8\},\{1,4,7\},\{1,5,6\},\{2,3,7\},\{2,4,6\},\{3,4,5\}.$$

There are $$7$$ such sets, and each gives $$3! = 6$$ ordered triples. Hence the required number of ordered triples is $$7\times6=42\;.$$

Answer : 42